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3 years ago

-4/5 : 1/3 I need help with this

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

Step-by-step explanation:

-4/5÷1/3= (-4/5)x(3/1)= -12/5=-2 2/5

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Write your answer in order of decreasing powers of x. (18x 5 + 6x 4 - 12x 3) ÷ 6x 2

miv72 [106K]

$\frac{18x^5 +6x^4 - 12x^3}{6x^2} = \frac{(6x^2)(3x^3 +x^2 - 2x)}{6x^2} = 3x^3 +x^2 - 2x$

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3 years ago

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MERRY CHRISTMASS!!! lets end the year by answering this math question

Andre45 [30]

Answer:

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3 years ago

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Simply (tan^2 x - sec^2 x)(sin^2 x + cos^2 x)

kati45 [8]

we can use pythagorean identieis

remember that $sin^2(x)+cos^2(x)=1$

also remember that $tan(x)=\frac{sin(x)}{cos{x}}$ and $sec(x)=\frac{1}{cos(x)}$

so if we take $sin^2(x)+cos^2(x)=1$ and divide both sides by $cos^2(x)$ we get

$\frac{sin^2(x)}{cos^2(x)}+\frac{cos^2(x)}{cos^2(x)}=\frac{1}{cos^2(x)}$

$(\frac{sin(x)}{cos(x)})^2+1=(\frac{1}{cos(x)})^2$

$tan^2(x)+1=sec^2(x)$

subtracting $1+sec^2(x)$ from both sides

$tan^2(x)-sec^2(x)=-1$

now subsitute into original problem

$\frac{tan^2(x)-sec^2(x)}{sin^2(x)+cos^2(x)}=$

$\frac{-1}{1}=$

$-1$

the answer is -1

6 0

3 years ago

Solve the equation 4 sin (x +30°) = 2 for 0 < x < 365

tatuchka [14]

$4 \sin(x + 30) = 2 \\ \sin(x + 30 ) = \frac{2}{4} \\ \sin(x + 30) = \frac{1}{2} \\ \sin(x + 30) = \sin(30) \\ x + 30 = 30 \\ x = 30 - 30 \\ x = 0$

Hence, x = 0°, 180° and 360° Ans

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3 years ago

While grading her students' most recent quiz on equation solving, Mrs. Jones calculated that approximately forty percent of her

EastWind [94]

The given equation is,

$3(-n+4)+5n=2n$

Part 1:

Solve the above equation for n.

Using distributive property of multiplication,

$\begin{gathered} -3n+3\times4+5n=2n \\ -3n+12+5n=2n \end{gathered}$

Now, group the like terms.

$-3n+5n-2n+12=0$

Now, add the like terms.

$\begin{gathered} -5n+5n+12=0 \\ 0=-12 \end{gathered}$

0=-12 is a false statement.

Since we obtained a false statement, the given equation has no solution.

So, option b is correct.

Part 2:

The option (a) is n=3.

If while solving an equation, a single value is obtained for the variable, then the equation has only a single solution. If n=3 is obtained after solving the equation, then the student should chose option (a) as the answer.

The option (b) is "no solution".

While solving an equation, if we obtain an equation which is mathematically false, then the equation will have no solution. So, no solution is chosen when no value is obtained for n and the final equation is mathematically false.

The option (c) is "infinitely many solutions".

While solving an equation, if we obtain an equation which is mathematically correct such as 0=0, 7=7 etc., then the equation will have infinietly many solutions. So, infinitely many solutions is chosen when no value is obtained for n and the final equation is mathematically correct.

8 0

1 year ago

-4/5 : 1/3 I need help with this​

-4/5 : 1/3 I need help with this